Circular convolution: Difference between revisions

Revision as of 16:11, 16 August 2021

Revision as of 17:02, 16 August 2021

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Then''':'''

{{NumBlk||<math>

:<math>\int_{t_o}^{t_o+T} h_{_T}(\tau)\cdot x_{_T}(t - \tau)\,d\tau = \int_{-\infty}^\infty h(\tau)\cdot x_{_T}(t - \tau)\,d\tau\ \triangleq\ (h *x_{_T})(t) = (x * h_{_T})(t).</math>|{{efn-uaEquationRef|Eq.1}}}}

|Proof:

:<math>\begin{align}

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:<math>\begin{align}

\int_{-\infty}^\infty h(\tau)\cdot x_{_T}(t - \tau)\,d\tau

&=\sum_{k=-\infty}^\infty \left[\int_{t_o+kT}^{t_o+(k+1)T} h(\tau)\cdot x_{_T}(t - \tau)\ d\tau\right] \quad t_0 \text{ is an arbitrary parameter}\\

&=\sum_{k=-\infty}^\infty \left[\int_{t_o}^{t_o+T} h(\tauu + kT)\cdot \underbrace{x_{_T}(t - \tauu-kT)}_{x_{_T}(t - \tauu), \text{ by periodicity}}\ d\taudu\right] \quad \text{substituting } u\tau \rightarrowtriangleq \tau+-kT\\

&=\int_{t_o}^{t_o+T} \left[\sum_{k=-\infty}^\infty h(\tauu + kT)\cdot x_{_T}(t - \tauu)\right]\ d\tau\\

&=\int_{t_o}^{t_o+T} \underbrace{\left[\sum_{k=-\infty}^\infty h(\tauu + kT)\right]}_{\triangleq \ h_{_T}(\tauu)}\cdot x_{_T}(t - \tauu)\ ddu\\tau

&=\int_{t_o}^{t_o+T} h_{_T}(\tau)\cdot x_{_T}(t - \tau)\ d\tau \quad \text{substituting } \tau \triangleq u

\end{align}</math>

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}}

Both forms can be called ''periodic convolution''.{{efn-la

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*[[Circulant matrix]]

*[[Hilbert transform#Discrete Hilbert transform|Discrete Hilbert transform]]

== Notes ==

{{notelist-ua}}

== Page citations ==